If M, N are integers greater than 1 such that 2M<N, which of the foll

Hi, this is my first answer. Kindly check and suggest if my approach is right.

Given, 2M<N => Assuming the AP has only 2 numbers in the series, M and N, M+N>3M.
Thus solving 3M with the options (54/2, 52/2, 50/2) should give a reminder, suggesting that the sum of M+N is greater than the values.

That is,
3M=27, No remainder. Therefore, M+N not greater than 3M.
3M=26, Remainder = 2. Therefore, M+N>3M.
3M=25, Remainder = 1. Therefore, M+N>3M.

Hence, (ii) and (iii) could be the answers, hence D.

Hi OhsostudiousMJ ,

Welcome to our GMAT Club community!

You have discussed whether M+N is greater than 3M, but this is not our focus. This problem is hard, even in terms of understanding what is going on...

Suggestion: start dealing with easier questions (600-700 or below 600). You can find each question´s level looking at the questions tags when they are posted.

Regards and success in your studies!
Fabio.

Although the problem seems hard, the focused-numbers (half the numbers given) are SMALL... and that´s the hint to try the "organized manual work" (chosen in previous solutions)!

We are looking for the 25 (III), 26 (II) and 27 (I) possibilities.

\({S_K} = 1 + 2 + \ldots + K = {{K\left( {K + 1} \right)} \over 2}\,\,\,\,\,\,\,\,\left( {{\rm{arithmetic}}\,\,{\rm{sequence}}} \right)\)

\({S_7} = 7 \cdot 4 = 28\,\,\, \Rightarrow \,\,\,\left\{ \matrix{\\
\,27 = 28 - 1 = \left( {1 + 2 + \ldots + 7} \right) - 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,2} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr \\
\,25 = 28 - 3 = \left( {1 + 2 + \ldots + 7} \right) - \left( {1 + 2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {N,M} \right) = \left( {7,3} \right)\,\,\,{\rm{with}}\,\,N > 2M\,\,\, \Rightarrow \,\,\,\,{\rm{viable!}} \hfill \cr} \right.\)

From the alternative choices given, we are sure we have found (EASILY!) the right answer!

The correct answer is (D).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.


POST-MORTEM:

Question 1.: how can we be sure that 26 (II) cannot be obtained? Please note that:

\(\left. \matrix{\\
{S_8} = 4 \cdot 9 = 36 \hfill \cr \\
{S_4} = 2 \cdot 10 = 10\,\,\, \hfill \cr} \right\}\,\,\, \Rightarrow \,\,\,26 = 36 - 10 = \left( {1 + 2 + \ldots + 7 + 8} \right) - \left( {1 + 2 + 3 + 4} \right)\)

and the ONLY reason (N,M) = (8,5) must be refuted is the fact that N>2M is false... What about other values for (N,M)? How can we be SURE there is not a single viable possibility?

Question 2.: which mathematical (GMAT-focused) properties could be useful to find all possible (N,M) pairs for a given LARGER value?

We will address both questions in our very next problem, here: https://gmatclub.com/forum/a-number-is- ... l#p2229235

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